Simplify the following expression and state the condition under which the simplification is valid. You can assume that $r \neq 0$. $n = \dfrac{-2}{2r(4r + 1)} \times \dfrac{16r^2 + 4r}{-10} $
Explanation: When multiplying fractions, we multiply the numerators and the denominators. $n = \dfrac{ -2 \times (16r^2 + 4r) } { 2r(4r + 1) \times -10 } $ $ n = \dfrac {-2 \times 4r(4r + 1)} {-10 \times 2r(4r + 1)} $ $ n = \dfrac{-8r(4r + 1)}{-20r(4r + 1)} $ We can cancel the $4r + 1$ so long as $4r + 1 \neq 0$ Therefore $r \neq -\dfrac{1}{4}$ $n = \dfrac{-8r \cancel{(4r + 1})}{-20r \cancel{(4r + 1)}} = -\dfrac{8r}{-20r} = \dfrac{2}{5} $